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Alexandrov's theorem, under Dependent Choice: a subspace of a complete metric space is completely metrizable exactly when it is
Statement
Assume Dependent Choice. For a subspace of a complete metric space , is completely metrizable if and only if is a subset of .
Facts & Assumptions
Given: The objects, hypotheses, and choice principles stated above.
If is a complete metric space and is in , then the subspace is completely metrizable. (Under the Axiom of Countable Choice, every subspace of a complete metric space is completely metrizable).
Assume Dependent Choice. If is a completely metrizable subspace of a metric space , then is a subset of . (Under Dependent Choice, every completely metrizable subspace of a metric space is ).
Proof
The empty subspace satisfies both conditions.
For a nonempty subspace apply the two preceding implications with the induced topology.
In the reverse direction use the given complete ambient metric; in the forward direction use only the existence of a compatible complete metric on the subspace, not completeness of the inherited metric.
The preceding construction and implications establish the assertion.
Depends on
Used by
Dependency tree · next 3 levels
Direct dependencies and their dependencies through the next three levels: 56 results over 14 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.
Sources
- David Marker, Descriptive Set Theory, §§1–2 (standard reference, not scraped)
- Michael Kunzinger, General Topology, §§11.3–11.4 (standard reference, not scraped)
- MFF General Topology course summary, §4.3 (standard reference, not scraped)
- Jesse Peterson, Real Analysis, §§3.6–3.7 (standard reference, not scraped)